1.

What is the [OH^(-)] in the final solution prepared by mixing 20.0ml of 0.50 M HCl with 30.0 ml of 0.10 M Ba(OH)_(2)

Answer»

0.10 M
0.40 M
0.0050 M
0.12 M

Solution :No. of milli equivalent of `HCl = 20 XX 0.05 = 1.0`
No. of milli equivalent of
`BA(OH)_(2) = 30 xx 0.10 xx 2 = 6.0`
After neutralization, no. of milli equivalents in 50 ML. of solution = (6-1) = 5
No. of milli equivalent of `OH^(-)` is 5 in 50 ml.
`[OH^(-)] = (5 xx 100)/(50) xx 10^(-3) = 0.1M`


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